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题目：
Given a digit string, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below.
Input:Digit string "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.
代码：
class Solution { public: vector<string> letterCombinations(string digits) { vector<string> ret; if ( digits.empty() ) return ret; map<int, string> digitLetters; digitLetters[0] = ""; digitLetters[1] = ""; digitLetters[2] = "abc"; digitLetters[3] = "def"; digitLetters[4] = "ghi"; digitLetters[5] = "jkl"; digitLetters[6] = "mno"; digitLetters[7] = "pqrs"; digitLetters[8] = "tuv"; digitLetters[9] = "wxyz"; vector<int> letterBegin(10,0); int index = 0; string tmp; Solution::combine(ret, tmp, index, digits, digitLetters, letterBegin); return ret; } static void combine( vector<string>& ret, string& tmp, int index, string& digits, map<int,string>& digitLetters, vector<int>& letterBegin) { if (index==digits.size()) { ret.push_back(tmp); return; } int curr = digits[index]-‘0‘; string letters = digitLetters[curr]; for ( int i = 0; i < letters.size(); ++i ) { letterBegin[curr] = i; tmp.push_back(letters[i]); Solution::combine(ret, tmp, index+1, digits, digitLetters, letterBegin); tmp.erase(tmp.end()-1); } } };
tips：
上述的代码是AC的。思路也就是常规dfs的思路：
1. 终止条件是层级到了digits的长度
2. 每一层递归相当于根据该位置的数字选一个字母
但是，个人感觉这道题的题干要求没说清楚；既然是letter combinations，而不是letter permutation，那么“ac”和“ca”就应该算一个combination，但是OJ之后发现默认“ac”和“ca”算两个。加入“ac”和“ca”算一个，有没有解法呢？代码如下：
class Solution { public: vector<string> letterCombinations(string digits) { vector<string> ret; if ( digits.empty() ) return ret; map<int, string> digitLetters; digitLetters[0] = ""; digitLetters[1] = ""; digitLetters[2] = "abc"; digitLetters[3] = "def"; digitLetters[4] = "ghi"; digitLetters[5] = "jkl"; digitLetters[6] = "mno"; digitLetters[7] = "pqrs"; digitLetters[8] = "tuv"; digitLetters[9] = "wxyz"; vector<int> letterBegin(10,0); int index = 0; string tmp; Solution::combine(ret, tmp, index, digits, digitLetters, letterBegin); return ret; } static void combine( vector<string>& ret, string& tmp, int index, string& digits, map<int,string>& digitLetters, vector<int>& letterBegin) { if (index==digits.size()) { ret.push_back(tmp); return; } int curr = digits[index]-‘0‘; string letters = digitLetters[curr]; for ( int i = letterBegin[curr]; i < letters.size(); ++i ) { letterBegin[curr] = i; tmp.push_back(letters[i]); Solution::combine(ret, tmp, index+1, digits, digitLetters, letterBegin); tmp.erase(tmp.end()-1); } } };
tips：
多维护一个letterBegin的数组，标记“当前的组合中，某个数字对应的字母序列中应该从第几个字母开始取”。
如果输入是“22”，那么给出的解集就是：
aa
ab
ac
bb
bc
cc
从这个例子可以看出来，算法的原理就是维护某一个数字对应字母序列：如果数字重复出现，那么对应的字母要保证字典序递增，这样就不会用重复的。
可惜题目并不是这么要求的。
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既然题目要求简单了，则再追求一个迭代的解法。AC的代码如下：
class Solution { public: vector<string> letterCombinations(string digits) { vector<string> ret; if (digits.empty()) return ret; ret.push_back(""); map<int, string> digitLetters; digitLetters[2] = "abc"; digitLetters[3] = "def"; digitLetters[4] = "ghi"; digitLetters[5] = "jkl"; digitLetters[6] = "mno"; digitLetters[7] = "pqrs"; digitLetters[8] = "tuv"; digitLetters[9] = "wxyz"; for ( size_t i = 0; i < digits.size(); ++i ) { int curr = digits[i]-‘0‘; string letters = digitLetters.find(curr)==digitLetters.end() ? "" : digitLetters[curr]; vector<string> tmp = ret; ret.clear(); for ( size_t j = 0; j < tmp.size(); ++j ) { for ( size_t k = 0; k < letters.size(); ++k ) { string ori = tmp[j]; ori += letters[k]; ret.push_back(ori); } } } return ret; } };
完毕。
【Letter Combinations of a Phone Number】cpp
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4522163.html
